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enyata [817]
3 years ago
6

How much work is done on 10.0 c of charge to move it through a potential diffrence of 9.0 v in 10.0s

Physics
1 answer:
GenaCL600 [577]3 years ago
4 0
We need more evidence to be provided
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An instrument that measures and detects vibrations in earth is known as a _______.
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A boom box, stereo set.
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Identify the volume units that are greater than one liter
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There are a lot of volume units, most specifically in English units, that are greater than one liter. The following are as follows:

gallon, which is equal to 4.54 liters
minim
barrel
cord
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bushel and;
hogshead

Also included are metric units which are dekaliter onwards.
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3 years ago
The Cosmological Argument for the existence of God intellectually follows cause-and-effect back in time -- in effect asking "Whe
ankoles [38]

Answer:

First uncaused cause

Explanation:

Aristotle states that an infinite regression in the principle of causality is not possible. If the regression were infinite, then there would never be a first cause (mover), since this would need another mover to start its motion. Therefore, according to Aristotle, there must be an unmoved mover that moves other things, but is not itself moved by any prior action,

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3 years ago
In which point the electric intensity of a sphere is maximum​
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Here, as the charge is uniformly distributed in the sphere, we will consider s as an area of the sphere which is, s=4πr2 and r is radius of the gaussian surface shown in the figure above. From this, it can be seen that
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3 years ago
Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun
Dominik [7]

Answer:

a. K_{Axis}=2.574x10^{29}J

b. K_{Orbit}=2.6577x10^{33}J

Explanation:

K_{Axis}=\frac{1}{2}I*w^2

I_{Sphere}=\frac{2}{5}*m*r^2

w=\frac{2\pi }{T} , T=24hrs*\frac{3600s}{1hr} =86400s

radius earth = 6371 km

mass earth = 5,972*10^24 kg

a.

K_{Axis}=\frac{1}{2}*\frac{2}{5}*m*r^2*(\frac{2\pi}{T})^2

K_{axis}=\frac{4\pi^2}{5}*5.98x10^{24}kg*(6.38x10^6m)^2*(\frac{1}{86400s})^2

K_{Axis}=2.574x10^{29}J

b.

T=1year*\frac{365day}{1year}*\frac{24hr}{1day}*\frac{3600s}{1hr}=31536000s

K_{Orbit}=\frac{1}{2}*I*w

I=m*r^2

K_{Orbit}=\frac{1}{2}*m*r^2*(\frac{2\pi}{T})^2

K_{Orbit}=\frac{4\pi^2}{5}*5.98x10^{24}*6.38x10^6m*(\frac{1}{31.536x10^6s})^2

K_{Orbit}=2.6577x10^{33}J

4 0
3 years ago
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